Question: $f(x) = -2x^{2}-4-g(x)$ $g(x) = -6x^{2}$ $ g(f(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = -2(1^{2})-4-g(1)$ To solve for the value of $f$ , we need to solve for the value of $g(1)$ $g(1) = -6(1^{2})$ $g(1) = -6$ That means $f(1) = -2(1^{2})-4-(-6)$ $f(1) = 0$ Now we know that $f(1) = 0$ . Let's solve for $g(f(1))$ , which is $g(0)$ $g(0) = -6(0^{2})$ $g(0) = 0$